# How to find the remainder of ${289 \times 144^{25}}$ divided by ${71^{71}}$

I am solving this question.

Finding the remainder of

$$\frac{289\times 144^{25}}{71^{71}}$$

This is how I have tried solving it. First it can be simplified to $\frac{17^2 \times 2 ^{50} \times 3^{25}}{71^{71}}$. Now if we use Euler Totient rule we get for $\phi(71^{71}) = 71(1-\frac{1}{71}) = 70$.

For, $$\frac{a^{\phi(n)}}{n}$$

I would get remainder as $1$.

I can't proceed from this point. Can someone help me to find out the remainder and where I went wrong?

EDIT : My totient rule was wrong. It's $\phi(71^{71}) = 71^{71}(1-\frac{1}{71}) = 71^{70} \times 70$.

EDIT #2 : tried solving this in a different way.

$$\frac{289}{71} \times \frac{144^{25}}{71^{70}}$$

For the first separated division we get $289 \equiv 5 \bmod 71$. Now further simplifying the second, $$\frac{12^{50}}{71^{70}} = \left( \frac{12^5}{71^7} \right)^{10} = \left( \frac{12}{71} \right)^{70} \times \frac{1}{12^{20}}$$

Now I get $12 \equiv -59 \bmod 71$. I'm stuck again here.

• Your Totient formula use is off. For a prime power $p^k$ it should be $p^k(1-1/p)=p^k-p^{k-1}.$ However how does that help in doing the problem? – coffeemath Dec 2 '16 at 13:42
• $$\phi(71^{71})= = 71^{71}(1-\frac{1}{71}) = 70 \cdot 71^{70}$$Use Hansel Lemma – N. S. Dec 2 '16 at 13:42
• The numerator is smaller than the denominator, so you (just) have to compute the numerator and you can call it a day (or a month) – Ross Millikan Dec 2 '16 at 14:58

${289 \times 144^{25}} \approx 2.6 \times 10^{56} < 2.8 \times 10^{131} \approx 71^{71}$. So the remainder is ${289 \times 144^{25}}$.
$17^2 12^{25} 12^{25} < 71^2 71^{25} 71^{25}$
• Should that be $17^2 12^{25} 12^{25}$ on the left-hand side? – hmakholm left over Monica Dec 2 '16 at 13:49
• I didn't understand the hint properly. Also I tried solving this way, $289 ≡ 5$ mod $71$ $and$ 144 ≡ -2 $mod$ 71$– Ankit Panda Dec 2 '16 at 14:12 • @sudoankit:$x \bmod y = x $whenever$0\le x <y\$. – hmakholm left over Monica Dec 2 '16 at 14:15